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Find the Height of a Cliff

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    You want to know the height of a cliff so you throw a rock off the edge with initial speed v0 = 42 m/s upwards, inclined at an angle θ = 27° with respect to the horizontal. You have a friend that records the time that it takes for the rock to hit the bottom of the canyon below. It takes 7.6 s for the rock to hit the bottom of the canyon from when you throw it.


    2. Relevant equations

    v0x = v0*cos(theta)
    v0y = v0*sin(theta)

    ay = -g
    ax = 0

    vy(t) = v0*sin(theta) - g*t
    vx(t) = v0*cos(theta)

    x(t) = h + v0*sin(theta)*t - 1/2*g*t^2
    y(t) = v0*cos(theta)*t

    h = height of the cliff
    v = velocity
    v0 = initial velocity
    a = acceleration

    3. The attempt at a solution

    I am just confused about how to find the height of the cliff. I can't use the initial time because the initial time is zero. I tried solving for the x-distance at the final time. but I don't know how that would help. Any suggestions please?!?!
     
  2. jcsd
  3. Sep 16, 2012 #2
    You have your equations for the x and y position mixed up. You only need to worry about what's happening in the vertical (y) direction.

    The position equation for the y direction is y = h + v0*sin(theta)*t - 1/2*g*t^2

    In this equation, y = 0 because the rock will land on the ground level which is easiest to assume is zero.

    t = 7.6s

    Plug in everything else and it should work out to get you an answer of 138.6 meters (give or take).
     
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